To be more precise, I am interested in knowing if the intuition that a Euclidean zero vector does not have a particular direction is actually correct, and if there is a rigorous formulation that would back it up.

Wikipedia's entries on the zero vector seems to agree with that intuition, but as always, one shouldn't blindly trust Wikipedia: in one place, it is stated that a zero vector "is orthogonal to all other vectors with the same number of components," while in another, it is stated that "two vectors can be considered orthogonal if and only if their dot product is zero, and they have non-null length." Correct me if I am wrong, but these two statements contradict each other.

This question popped up in my head when I heard someone arguing that since a Euclidean vector is defined as a geometric entity that has both a magnitude and direction, and since a zero vector is a vector with length 0, then it is only fitting for a zero vector to have a "direction 0." I am personally inclined to say that a zero vector does not have a particular direction, but as I said, I would like to know if there is a rigorous formulation that would lead to this conclusion.

And to put the question in a less "discussion-inducing" form, is there an agreement on the direction of a Euclidean zero vector?

6 Answers
6

The zero vector has no particular direction; this is consistent with the fact that it is orthogonal to every other vector. (It doesn't really make sense to say it has "direction 0", since direction is not a magnitude; "direction 0" makes no more sense than "direction 1"
or "direction 5.873".)

Alternatively, you could say that it points in every direction, but with zero magnitude,
since if you take any vector and multiply it by zero, you get the zero vector. "Every
direction" is the same as "no particular direction"; it's just a different way of phrasing
things.

This is one flaw with the traditional description of a vector as being a pair consisting of
a magnitude and a direction: For the zero vector, the magnitude is zero, but the direction
is arbitrary.

I like thinking of it as being consistent with the convention that the argument (phase angle) of 0 is indeterminate.
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Guess who it is.Aug 31 '10 at 0:25

1

I think it comes down to direction is really just an analogy. Direction is a consequence of the vector having components and us drawing it a particular way on 4-square paper - but as soon as you get to 4 dimensional vectors, you can no longer use the "direction" analogy.
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boboboboNov 19 '11 at 17:52

As Matt E points out in his answer, the "direction" of a vector $v \in \mathbb{R}^n$ is not a number. What then is it? A natural answer is that a direction is an element of the unit sphere $S^{n-1} = {(x_1,\ldots,x_n) \in \mathbb{R}^n \ | \ x_1^2 + \ldots + x_n^2 = 1 }$.

Then the map, say $D$, which assigns each $v \in \mathbb{R}^n \setminus {0}$ to its direction to its direction $D(v) = \frac{v}{||v||} \in S^{n-1}$ is a deformation retraction. It follows that $\mathbb{R}^n \setminus {0}$ has the same homotopy type as the sphere $S^{n-1}$. In particular, $\mathbb{R}^n \setminus {0}$ does not have the same homotopy as $\mathbb{R}^n$ -- i.e., it is not contractible. This can be taken as a precision of the idea that there is no natural way to extend $D$ so as to be defined at the zero vector. (For that matter, there is evidently no continuous extension of $D$ to $\mathbb{R}^n$, since $D$ is surjective on any deleted neighborhood of $0$.)

Note also that any algebraic geometer will inevitably be reminded of projective space, in which the zero vector must be excluded for similar reasons. In this context, (real) projective $n-1$-space is obtained by failing to distinguish between the directions of $v$ and $-v$. Topologically, this amounts to taking a quotient of $S^{n-1}$ by identifying antipodal points.

How about this one: the direction of a vector $x$ is a vector $u$ of unit length (that is, $|u| = 1$) such that $cx = u$ for some positive real number $c$.

According to the definition, it is clear that if $x = 0$, then there exists no $c$ such that $cx$ is a unit vector.

Also, it is clear that the direction $d(x)$ of a vector $x$ is a function mapping $x$ to a unit vector. For any given $x$, the coefficient $c$ is just $\frac{1}{|x|}$, and so the function is this: $$d(x) = \frac{x}{|x|}$$.

It is clear that the function is not defined at $x = 0$, where it invokes division by zero.

I think it is possible for a zero magnitude vector to have direction. Consider a particle moving in a circle at constant speed. Velocity is variable because direction changes. magnitude of acceleration is zero because speed is constant. Acceleration (directed towards center) is variable because direction changes every instant. Clearly acceleration is present with magnitude zero.

The magnitude of the acceleration is not zero. In your case it is $\frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius of the circle.
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user50229Mar 20 '13 at 16:04

@anjumrashid: Welcome to MSE! You might want to consider updating or delteing your response. Its happened to me many times and will many more! Regards
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AmzotiMar 21 '13 at 12:19

You contradict yourself plainly in the last two sentences. First you claim that acceleration is directed toward the center and that it is variable. Then you say that it has magnitude zero? How can something variable have magnitude zero? You're missing the fact that the acceleration is always at right angles to the velocity. Yes, angular velocity is constant and so angular acceleration is zero.
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KazMar 27 '13 at 20:03

when we take cross product of two vectors which are perpendicular, we get a vector having magnitude zero. But we know the direction of this vector by using right hand thumb rule or screw rule. so we know the direction of a null vector...! what do you say about this?

Zero is the cross product of two parallel vectors, not two perpendicular (non-zero) vectors. And with parallel vectors, the right-hand rule gives no particular direction.
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Andreas BlassMay 17 '13 at 15:57